Symmetric Bending of Beams


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1 Symmetric Bending of Beams beam is any long structural member on which loads act perpendicular to the longitudinal axis. Learning objectives Understand the theory, its limitations and its applications for strength based design and analysis of symmetric bending of beams. Develop the discipline to visualize the normal and shear stresses in symmetric bending of beams. 61
2 6.5 Due to the action of the external moment M z and force P, the rigid plate shown in Fig. P6.5 was observed to rotate by 2 o and the normal strain in bar 1 was found ε 1 = 2000 µ in in. Both bars have an area of crosssection of = 1/2 in 2 and a Modulus of Elasticity of E = 30,000 ksi. Determine the applied moment M z and force P. z y x P 2 in Bar 2 Bar 1 4 in M z 48 in Fig. P
3 Internal Bending Moment M z = yσ xx d σ d xx = 0 bove equations are independent of material model as these equations represents static equivalency between the normal stress on the entire crosssection and the internal moment. The line on the crosssection where the bending normal stress is zero is called the neutral axis. Location of neutral axis is chosen to satisfy σ d xx = 0. Origin of y is always at the neutral axis, irrespective of the material model. 63
4 6.8 Steel (E steel = 30,000 ksi) strips are securely attached to a wooden (E wood = 2,000 ksi) beam as shown below. The normal strain at the crosssection due to bending about the zaxis is ε xx = 100 y µ where y is measured in inches, and the dimensions of the crosssection are d =2 in, h W =4 in and h S = (1/8) in. Determine the equivalent internal moment M z. y d Steel h S z Wood h W Steel h S 64
5 Theory of symmetric bending of beams Limitations The length of the member is significantly greater then the greatest dimension in the crosssection. We are away from the regions of stress concentration. The variation of external loads or changes in the crosssectional areas are gradual except in regions of stress concentration. The crosssection has a plane of symmetry. The loads are in the plane of symmetry. The load direction does not change with deformation. The external loads are not functions of time. 65
6 Theory objectives: To obtain a formula for the bending normal stress σ xx, and bending shear stress τ xy in terms of the internal moment M z and the internal shear force V y. To obtain a formula for calculation of the beam deflection v(x). The distributed force p(x), has units of force per unit length, and is considered positive in the positive ydirection. Displacements Kinematics 1 External Forces and Moments Strains 4 Equilibrium Internal Forces and Moments Static Equivalency 3 Stresses Material Models 2 66
7 Kinematics z y x Original Grid Deformed Grid ssumption 1 ssumption 2 Squashing, i.e., dimensional changes in the ydirection, is significantly smaller than bending. v ε yy = 0 y v = v ( x ) Plane sections before deformation remain plane after deformation. u = u o ψ y y x ψ u o ssumption 3 Plane perpendicular to the beam axis remain nearly perpendicular after deformation. γ xy
8 ssumption 4 B 1 B 2 O C ψ ψ u D 1 D R R  y y Strains are small. B y 1 D 1 ψ bending normal strain ε xx varies linearly with y and has maximum value at either the top or the bottom of the beam. 2 1 d v  = ( x) is the curvature of the deformed beam and R is the radius R dx 2 of curvature of the deformed beam. Material Model ssumption 5 Material is isotropic. ssumption 6 Material is linearly elastic. ssumption 7 There are no inelastic strains. From Hooke s Law: σ xx tanψ ψ Method I = dv dx B = CD = CD 1 B 1 B ε xx = = B ε xx = Method II y  R ( R y) ψ R ψ R ψ u = ysin ψ y ψ 2 u ε xx dψ d v = lim = y = y ( x) x 0 x dx dx 2 2 d v ε xx = y ( x) dx 2 2 d v = Eε xx, we obtainσ xx = Ey ( x) dx 2 68
9 Location of neutral axis 2 σ d d v xx = 0 or Ey ( x) d or x 2 = 0 Eyd = 0 d ssumption 8 Material is homogenous across the crosssection of the beam. y d = 0 Neutral axis i.e, the origin, is at the centroid of the crosssection constructed from linearelastic, isotropic, homogenous material. The axial problem and bending problem are decoupled if the origin is at the centroid for linearelastic, isotropic, homogenous material bending normal stress σ xx varies linearly with y and is zero at the centroid. bending normal stress σ xx is maximum at a point farthest from the neutral axis (centroid). 69
10 6.20 The crosssection of a beam with a coordinate system that has an origin at the centroid C of the crosssection is shown. The normal strain at point due to bending about the zaxis, and the Modulus of Elasticity are as given. (a) Plot the stress distribution across the crosssection. (b) Determine the maximum bending normal stress in the crosssection. (c) Determine the equivalent internal bending moment M z by integration. ε xx = 200 µ E = 8000 ksi z 4 in y C 4 in 1 in 1 in 610
11 Sign convention for internal bending moment M z = yσ xx d The direction of positive internal moment M z on a free body diagram must be such that it puts a point in the positive y direction into compression. 611
12 Sign convention for internal shear force Internal Forces and Moment necessary for equilibrium Recall ssumption 3: Plane perpendicular to the beam axis remain nearly perpendicular after deformation. γ xy 0. From Hooke s Law: τ xy = Gγ xy Bending shear stress is small but not zero. Check on theory: The maximum bending normal stress σ xx in the beam should be nearly an order of magnitude greater than the maximum bending shear stress τ xy. V y = τ xy d The direction of positive internal shear force on a free body diagram is in the direction of positive shear stress on the surface. 612
13 6.26 beam and loading in three different coordinate system is shown. Determine the internal shear force and bending moment at the section containing point for the three cases shown using the sign convention described in Section kn/m 5 kn/m 5 kn/m x x x y 0.5 m 0.5 m 0.5 m 0.5 m 0.5 m y Case 1 Case 2 Case m y 613
14 Flexure Formulas σ xx = 2 d v Ey ( x) dx d v d v 2 M z = yσ xx d = y Ey ( x) d x 2 = ( x) Ey d d x 2 d For homogenous crosssections 2 d v Momentcurvature equation: M z = EI zz 2 d x I zz is the second area moment of inertia about zaxis. The quantity EI zz is called the bending rigidity of a beam crosssection. Flexure stress formula: σ xx = M z y I zz Two options for finding M z On a free body diagram M z is drawn as per the sign convention irrespective of the loading. positive values of stress σ xx are tensile negative values of σ xx are compressive. On a free body diagram M z is drawn at the imaginary cut in a direction to equilibrate the external loads. The tensile and compressive nature of σ xx must be determined by inspection. 614
15 6.20 The crosssection of a beam with a coordinate system that has an origin at the centroid C of the crosssection is shown. The normal strain at point due to bending about the zaxis, and the Modulus of Elasticity are as given. (d) Determine the equivalent internal bending moment M z by flexure formula. ε xx = 200 µ E = 8000 ksi z 4 in y C 4 in 1 in 1 in 615
16 Class Problem 1 The bending normal stress at point B is 15 ksi. (a) Determine the maximum bending normal stress on the crosssection. (b) What is the bending normal strain at point if E = 30,000 ksi. 616
17 6.15 Fig. P6.15(a) shows four separate wooden strips that bend independently about the neutral axis passing through the centroid of each strip. Fig. P6.15(b) shows the four strips are glued together and bend as a unit about the centroid of the glued crosssection. (a) Show that I G = 16I S, where I G is the area moment of inertias for the glued crosssection and I S is the total area moment of inertia of the four separate beams. (b) lso show, where σ G and σ S are the maximum bending normal stress at any crosssection for the glued and separate beams, respectively. (a) σ G = σ S 4 (b) Fig. P
18 6.30 For the beam and loading shown, draw an approximate deformed shape of the beam. By inspection determine whether the bending normal stress is tensile or compressive at points and B. M B Class Problem For the beam and loading shown, draw an approximate deformed shape of the beam. By inspection determine whether the bending normal stress is tensile or compressive at points and B. B 618
19 6.41 The beam, loading and the crosssection of the beam are as shown. Determine the bending normal stress at point and the maximum bending normal stress in the section containing point 619
20 6.46 wooden (E = 10 GPa) rectangular beam, loading and crosssection are as shown in Fig. P6.46. The normal strain at point in Fig. P6.46 was measured as ε xx = 600 µ. Determine the distributed force w that is acting on the beam. w kn/m 100 mm x 25mm 0.5 m 0.5 m Fig. P
21 Shear and Moment by Equilibrium Differential Beam Element Differential Equilibrium Equations: dv y dx The above equilibrium equations are applicable at all points on the beam except at points where there is a point (concentrated) force or point moment. Two Options for finding V y and M z as a function of x Integrate equilibrium equations and find integration constants by using boundary conditions or continuity conditions. This approach is preferred if p not uniform or linear. Make an imaginary cut at some location x, draw free body diagram and use static equilibrium equations to find V y and M z. Check results using the differential equilibrium equations above. This approach is preferred if p is uniform or linear. = p dm z dx = V y 621
22 6.51 (a) Write the equations for shear force and bending moments as a function of x for the entire beam. (b) Show your results satisfy the differential equilibrium equations. y 5 kn/m x 3 m 622
23 6.58 For the beam shown in Fig. P6.58, (a)write the shear force and moment equation as a function of x in segment CD and segment DE. (b) Show that your results satisfy the differential equilibrium equations. (c) What are the shear force and bending moment value just before and just after point D. Fig. P6.58 Class Problem 3 Write the shear force and moment equation as a function of x in segment B. 623
24 Shear and Moment Diagrams Shear and Moment diagrams are a plots of internal shear force V y and internal bending moment M z vs. x. Distributed force n integral represent area under the curve. To avoid subtracting positive areas and adding negative areas, define V = V y x 2 x 2 V 2 = V 1 + pdx M 2 = M 1 + Vdx x 1 x 1 y x (a) (b) (c) (d) w x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 w V 1 V 2 =V 1 w(x 2 x 1 ) V V 2 =V 1 +w(x 2 x 1 ) V V V x x 1 x 1 x 2 2 V V x 1 1 x 1 2 V y V y V V y y x x V 2 =V 1 w(x 2 x 1 ) 2 V V 1 2 =V 1 +w(x 2 x 1 ) 1 x Increasing incline of tangent 1 Decreasing incline of tangent Increasing incline of tangent Decreasing incline of tangent M M 2 M 2 1 M z M1 M z M 1 M z M M 1 z M 2 x 1 x 2 x 1 x 2 M 2 x 1 x 2 x 1 x 2 If V y is linear in an interval then M z will be a quadratic function in that interval. Curvature rule for quadratic M z curve. The curvature of the M z curve must be such that the incline of the tangent to the M z curve must increase (or decrease) as the magnitude of the V increases (or decreases). or The curvature of the moment curve is concave if p is positive, and convex if p is negative. 624
25 Point Force and Moments Internal shear force jumps by the value of the external force as one crosses the external force from left to right. Internal bending moment jumps by the value of the external moment as one crosses the external moment from left to right. Shear force & moment templates can be used to determine the direction of the jump in V and M z. template is a free body diagram of a small segment of a beam created by making an imaginary cut just before and just after the section where the a point external force or moment is applied. Shear Force Template V 1 V 2 M1 Moment Template M ext M 2 x x x x F ext V 2 = V 1 + F ext M 2 = M 1 + M ext Template Equations The jump in V is in the direction of F ext 625
26 6.70 Draw the shear and moment diagram and determine the values of maximum shear force V y and bending moment M z. M1 M ext M 2 x x M 2 = M 1 + M ext 626
27 6.77 Two pieces of lumber are glued together to form the beam shown Fig. P6.77. Determine the intensity w of the distributed load, if the maximum tensile bending normal stress in the glue limited to 800 psi (T) and maximum bending normal stress is wood is limited to 1200 psi. w lb / in 2 in 30 in 70 in 4 in 1 in Fig. P
28 Shear Stress in Thin Symmetric Beams Recollect problem 615 for motivation for gluing beams I G = 16I S σ G = σ S 4 Separate Beams Glued Beams Relative Sliding No Relative Sliding ssumption of plane section perpendicular to the axis remain perpendicular during bending requires the following limitation. Maximum bending shear stress must be an order of magnitude less than maximum bending normal stress. 628
29 Shear stress direction Shear Flow: q = τ xs t The units of shear flow q are force per unit length. The shear flow along the centerline of the crosssection is drawn in such a direction as to satisfy the following rules: the resultant force in the ydirection is in the same direction as V y. the resultant force in the zdirection is zero. it is symmetric about the yaxis. This requires shear flow will change direction as one crosses the yaxis on the centerline. 629
30 6.88 ssuming a positive shear force V y, (a) sketch the direction of the shear flow along the centerline on the thin crosssections shown.(b) t points, B, C, and D, determine if the stress component is τ xy or τ xz and if it is positive or negative. y B D z C Class Problem ssuming a positive shear force V y, (a) sketch the direction of the shear flow along the centerline on the thin crosssections shown.(b) t points, B, C, and D, determine if the stress component is τ xy or τ xz and if it is positive or negative. B y D z C 630
31 Bending Shear Stress Formula ( N s + d N s ) N s s is the area between the free surface and the point where shear stress is being evaluated. Define: ssumption 9 + τ sx tdx = 0 τ sx t d τ sx t σ dx d d = xx = d dx = Q z s = τ sx t s = yd s The beam is not tapered. M z y d dx I zz M z Q z I zz = dn s dx d dx M z yd I zz s Q z dm q tτ sx z Q z V y V y Q = = = z τ I zz dx I zz sx = τ xs = I zz t 631
32 Calculation of Q z Q z = yd s is the area between the free surface and the point where shear stress is being evaluated. Q z is zero at the top surface as the enclosed area s is zero. Q z is zero at the bottom surface ( s =) by definition of centroid. s z s Line along which Shear stress is being found. Centroid of 2 y Centroid of s Q z s y s y s Neutral xis y 2 Q z = 2 y 2 = 2 Q z is maximum at the neutral axis. Bending shear stress at a section is maximum at the neutral axis. 632
33 Bending stresses and strains Top or Bottom Neutral xis Point in Web Point in Flange ε xx γ xy σ xx = ε νσ xx E yy νσ xx = = νε E xx ε zz = = νε E xx τ xy τ = γ xz G xz = G 633
34 6.97 For the beam, loading and crosssection shown, determine: (a) the magnitude of the maximum bending normal and shear stress. (b) the bending normal stress and the bending shear stress at point. Point is on the crosssection 2 m from the right end. Show your result on a stress cube. The area moment of inertia for the beam was calculated to be I zz = 453 (10 6 ) mm
35 Shear Flow q = V y Q z I zz s s V F s V F P q = V F s q = 2V F s 635
36 6.104 wooden cantilever box beam is to be constructed by nailing four 1 inch x 6 inch pieces of lumber in one of the two ways shown. The allowable bending normal and shear stress in the wood are 750 psi and 150 psi, respectively. The maximum force that the nail can support is 100 lbs. Determine the maximum value of load P to the nearest pound, the spacing of the nails to the nearest half inch, and the preferred nailing method.* Joining Method 1 Joining Method
37 6.115 cantilever, hollowcircular aluminum beam of 5 feet length is to support a load of 1200lbs. The inner radius of the beam is 1 inch. If the maximum bending normal stress is to be limited to 10 ksi, determine the minimum outer radius of the beam to the nearest 1/16th of an inch. 637
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